Question Video: Solving Problems Involving Permutations and Combinations | Nagwa Question Video: Solving Problems Involving Permutations and Combinations | Nagwa

# Question Video: Solving Problems Involving Permutations and Combinations Mathematics • Third Year of Secondary School

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The number of different ways 5 persons can sit in 5 seats in the form of a row equals ＿. [A] 5 × 5 [B] 5 × 4 × 3 × 2 × 1 [C] 5 + 5 [D] 1

02:53

### Video Transcript

The number of different ways five persons can sit in five seats in the form of a row equals what. Five multiplied by five, five multiplied by four multiplied by three multiplied by two multiplied by one, five plus five, or one.

So we have five people, and we want to think about the number of different ways they can sit in five seats. Here are those five seats. Now each of these different orders that the five people can sit in is unique. For example, here’s one possible order. Or they could choose to sit like this. Our job is to work out how many different unique orderings there are.

Looking at the four options we’ve been given, we can see that this question is more interested in working out what the calculation is as opposed to the actual numerical answer. So our approach is going to be to think about the logic for how we would work this out. Let’s take those five empty seats and think about the number of possibilities for each seat. In the first seat of the row, any of the five people could sit here. So there are five possibilities. Once that seat is full, there are four people remaining and four seats. So there are four possibilities for who sits in the next seat.

Now there are three seats remaining and three people. So there are three possibilities for who sits in the third seat. In the same way, there are two possibilities for who sits in the next seat. And by the time we get to the last seat, only one person is left. So there’s only one choice. Any of the five possibilities for the first seat can be combined with any of the four possibilities for the next seat, which can be combined with any of the three possibilities for the next seat, which can be combined with any of the two possibilities for the next seat, and so on.

The calculation that we’re looking for then is five multiplied by four multiplied by three multiplied by two multiplied by one. That’s this option here. Of course, multiplying by one doesn’t actually have any effect, but we’ll include it for completeness.

You may be familiar with another way of writing this calculation. Five multiplied by four multiplied by three multiplied by two multiplied by one can be written as five and then an exclamation point, which stands for five factorial. This means the product of all the integers from one up to five. In fact, this question illustrates a general rule, which is that if we have 𝑛 unique objects, then there are 𝑛 factorial possible orderings of those 𝑛 objects. That’s 𝑛 multiplied by 𝑛 minus one multiplied by 𝑛 minus two all the way done to one.

We’ve shown that the number of different ways these five people can sit in a row of five seats can be calculated as five multiplied by four multiplied by three multiplied by two multiplied by one.

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